Environmental Engineering Reference
In-Depth Information
which corresponds to a power of 4
10
13
kW. The power associated with passage of
electric current through the atmosphere is
UI
D
5
10
5
kW, where
U
D
300 kV
is the Earth's potential and
I
1700 A is the atmospheric current. Since the cloud
potential during a thunderstorm exceeds the potential of the Earth by a factor of
10
3
, we find that the power expended in the charging of the Earth is three orders
of magnitude greater than the power of discharging of the Earth, but five orders of
magnitude smaller than the power consumed in the evaporation of water. There-
fore, the power expended on electric processes during the circulation of water in
the atmosphere is relatively small. To create the observed charging current, it is
necessary that the charge transfer should be 1.4
D
10
10
C/g of water. We will use
this value for subsequent estimates.
Being guided by micrometer-sized water drops, we consider the case (6.5) where
the drop radius
r
0
is large compared with the mean free path of ions in air, which
gives
r
0
m for atmospheric air. Then the average charge
Z
of a drop (an
aerosol particle) that is established by attachment of negative and positive ions to
an aerosol is given by (6.12)
0.1
μ
r
0
T
e
2
ln
K
K
C
j
Z
jD
,
(6.104)
where
K
and
K
C
arethemobilitiesofthenegativeandpositiveionsinatmo-
spheric air. The mobility of various ions in nitrogen, the basic component of air,
are given in Table 6.10. Comparison of the mobilities for positive and negative ions
shows that they are nearby. From this it follows that in principle aerosol particles
in the Earth's atmosphere may be charged both negatively and positively, although
the negative charge of aerosol particles is preferable. On the basis of the charge of
clouds, one can conclude that the negative charge of aerosol particles is realized in
roughly 90% of cases.
Let us consider a cloud where the mobility of negative molecular ions is higher
than that of positive ions, and we take for definiteness
K
/
K
C
D
1.1. We assume
that water drops have a spherical shape with radius
r
0
and extract the dependence
of various parameters on the drop radius, and reduce various parameters to the
drop radius
a
m. Then we have for the mass of the drop
m
expressed in
grams,
the
fall velocity
w
of a water drop expressed in cm/s, and the average drop
charge
D
1
μ
j
Z
j
at temperature
T
D
300 K given in electron charges
10
12
r
0
a
0.012
r
0
a
1.7
r
0
a
.
3
2
m
D
4.2
,
w
D
,
j
Z
jD
These formulas give for the ratio of the drop charge to its mass (expressed in C/g)
10
8
r
0
a
j
m
D
Z
2
6.6
.
Since the ratio of the total electric charge transferred in the Earth's atmosphere to
the total water mass is 1.4
10
10
C/g, we obtain from this that the aerosol size is
